Scenario Recall For This Week's Discussion You Considered

Scenariorecall That For This Weeks Discussion You Considered Data Re

Recall that for this week’s Discussion you considered data related to opening or attracting a new restaurant. Now consider that you ask 20 participants to estimate how many times a month they go out to dinner and you receive these responses: 1, 2, 5, 8, 2, 4, 8, 4, 2, 3, 6, 8, 7, 5, 8, 4, 0, 7, 6, and 18. Assignment: To complete this assignment, submit by Day 7 calculations of the following measures of central tendency and variability using the data set provided. Include an explanation of how you calculated each measure and what information each measure gives you about the dining behavior of the sample. Finally, create a data file in SPSS and run analyses to find the mean and standard deviation.

Note: Your hand-calculated mean and standard deviation will differ somewhat from the calculations in SPSS due to rounding. Hand-calculated mean: SPSS mean: Median: Mode: Range: Deviation of the highest score from the mean: Hand-calculated standard deviation (Please also state the hand-calculated values for ΣX2 and (ΣX)2): SPSS standard deviation: Explain how the standard deviation (SD) and the deviation of a single score differ in the information they provide. Explain how each measure (mean, median, mode, deviation of the highest score from the mean, and standard deviation) would change if the score of 18 was eliminated from the data set. Explain the type of distribution (positive skew, negative skew, bimodal distribution, or normal distribution) your data create. Explain how you know the type of distribution and what the data tells you about your sample. Submit three documents for grading: your text (Word) document with your answers and explanations to the application questions, your SPSS Data file, and your SPSS Output file. Resources: Readings Heiman, G. (2015). Behavioral sciences STAT 2 (2nd ed). Stamford, CT: Cengage. Review Section 2-3 “Types of Frequency Distributions” (pp.25-28) Chapter 3, “Summarizing Scores with Measures of Central Tendency” (pp.36-49) Chapter 4, “Summarizing Scores with Measure of Variability” (pp.52-65) Chapter 1 Review Card (p. 1.4) Chapter 3 Review Card (p. 3.. A bank randomly selected 247 checking account customers and found that 107 of them also had savings accounts at this same bank. Construct a 90% confidence interval for the true proportion of checking account customers who also have savings accounts. (Round your answers to three decimal places.)

Paper For Above instruction

This assignment involves comprehensive statistical analysis of a data set derived from participants' self-reported frequencies of dining out per month. The process includes calculation of measures of central tendency and variability, interpretation of these measures, creating and analyzing data in SPSS, and understanding the distribution characteristics of the data. Additionally, it involves constructing a confidence interval for a proportion related to banking data, which broadens the scope to inferential statistics. This essay provides detailed explanations of calculations, their significance, and practical implications for understanding customer behavior and sample distributions.

Calculations of Measures of Central Tendency and Variability

The data points provided by 20 participants are: 1, 2, 5, 8, 2, 4, 8, 4, 2, 3, 6, 8, 7, 5, 8, 4, 0, 7, 6, and 18. To analyze this data, first we calculate the mean, median, mode, range, and standard deviation both by hand and via SPSS.

The hand-calculated mean is derived by summing all values and dividing by the number of data points:

Sum (ΣX) = 1 + 2 + 5 + 8 + 2 + 4 + 8 + 4 + 2 + 3 + 6 + 8 + 7 + 5 + 8 + 4 + 0 + 7 + 6 + 18 = 125

Mean = ΣX / N = 125 / 20 = 6.25

In SPSS, the software would similarly compute the mean, which should approximate 6.25, with minor differences due to rounding.

The median, the middle value when data are ordered, is obtained by sorting the data: 0, 1, 2, 2, 2, 3, 4, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 8, 8, 18. The middle two values (since N=20) are the 10th and 11th data points: 5 and 5. Therefore, the median is (5+5)/2 = 5.

The mode, the most frequently occurring value, appears to be 8, which occurs 4 times.

The range, the difference between the maximum and minimum values, is 18 - 0 = 18.

The deviation of the highest score from the mean is 18 - 6.25 = 11.75, indicating how far the maximum value is from the central tendency.

Calculating the hand-standard deviation involves determining ΣX² and (ΣX)²:

ΣX² = (1)² + (2)² + (5)² + (8)² + (2)² + (4)² + (8)² + (4)² + (2)² + (3)² + (6)² + (8)² + (7)² + (5)² + (8)² + (4)² + (0)² + (7)² + (6)² + (18)² = 0 + 1 + 4 + 25 + 64 + 4 + 16 + 64 + 16 + 9 + 36 + 64 + 49 + 25 + 64 + 16 + 0 + 49 + 36 + 324 = 1032

(ΣX)² = (125)² = 15,625

Using the formula for the sample standard deviation:

s = sqrt[ (ΣX² - (ΣX)² / N) / (N - 1) ]

Plugging in the numbers:

s = sqrt[ (1032 - 15,625/20) / 19 ] = sqrt[ (1032 - 781.25) / 19 ] = sqrt[ 250.75 / 19 ] ≈ sqrt[13.21] ≈ 3.63

The SPSS standard deviation should be very close to 3.63, with minor differences due to rounding.

Interpretation of Standard Deviation and Deviations of Scores

The standard deviation (SD) quantifies the average amount its data points deviate from the mean, offering insight into data variability. The deviation of a single score (like the highest score from the mean) indicates how far a specific data point is from the mean, which helps identify outliers or extreme values. While SD provides a global measure of variability across all data, the deviation of a single score pinpoints the extremity of a specific observation.

Effects of Removing the Score of 18

If the score of 18 were eliminated, the data set would have 19 data points, and the calculations would change. The new sum would be 125 - 18 = 107, resulting in a new mean of 107 / 19 ≈ 5.63. The highest score would then be 8, so the deviation of the highest score from the mean would become 8 - 5.63 ≈ 2.37, significantly smaller than before. Additionally, the variability (standard deviation) would likely decrease, reflecting less dispersion. The median might shift slightly, and the mode would probably remain at 8 if it still occurs most frequently.

Distribution Shape and Characteristics

By examining the data's skewness and histogram, the distribution appears positively skewed (right-skewed), primarily due to the outlier value of 18, which extends the right tail. The presence of a mode at 8, which occurs most frequently, and symmetrical data around the median suggest a skewed distribution. The positive skew indicates that most participants go out 0-8 times a month, but a few go significantly more often, skewing the distribution to the right.

This skewness can be confirmed by calculating skewness coefficients, and visualizing the histogram or boxplot would illustrate the skewness more clearly. The skewed distribution suggests that while most participants dine out moderately frequently, a minority has very high dining-out frequencies, influencing the overall data pattern.

Constructing Confidence Interval for Proportion

The second part of the assignment involves constructing a 90% confidence interval for the proportion of bank customers who also have savings accounts. Given that 107 out of 247 checking account customers also had savings accounts, the sample proportion (p̂) is:

p̂ = 107 / 247 ≈ 0.433

The formula for a confidence interval for a proportion is:

p̂ ± Z * sqrt( p̂(1 - p̂) / n )

At 90% confidence, the Z-score is approximately 1.645. Calculating the standard error (SE):

SE = sqrt( 0.433 (1 - 0.433) / 247 ) ≈ sqrt( 0.433 0.567 / 247 ) ≈ sqrt( 0.2457 / 247 ) ≈ sqrt(0.000994) ≈ 0.0315

The margin of error (ME) is:

ME = 1.645 * 0.0315 ≈ 0.0519

Thus, the confidence interval is:

[0.433 - 0.0519, 0.433 + 0.0519] ≈ [0.381, 0.485]

Rounding to three decimal places, the 90% confidence interval is (0.381, 0.485). This interval suggests that between 38.1% and 48.5% of all checking account customers at this bank also have savings accounts, with 90% confidence.

Conclusion

This comprehensive analysis illustrates how descriptive statistics like mean, median, mode, and range provide insights into the central tendency and variability of individual dining behaviors, while inferential statistics allow us to estimate population parameters based on sample data. Understanding the distribution shape helps interpret data skewness and outliers, which influence decision-making in restaurant marketing strategies and banking services. Using software like SPSS enhances the precision and efficiency of these statistical calculations, supporting data-driven decisions tailored to customer behaviors and financial patterns.

References

  • Heiman, G. (2015). Behavioral sciences STAT 2 (2nd ed.). Stamford, CT: Cengage.
  • Field, A. (2013). Discovering statistics using IBM SPSS statistics. Sage.
  • Laerd Statistics. (2017). Confidence interval for a population proportion (p). https://statistics.laerd.com/statistical-guides/confidence-interval-for-proportion.php
  • Tabachnick, B. G., & Fidell, L. S. (2013). Using multivariate statistics. Pearson.
  • Warner, R. (2013). Applied statistics: From bivariate through multivariate techniques. Sage.
  • Levine, D. M., & Stephan, D. (2011). Statistics for managers using Microsoft Excel. Pearson.
  • Agresti, A., & Franklin, C. (2017). Statistics: The art and science of data. Pearson.
  • Ghasemi, A., & Zahediasl, S. (2012). Normality tests for statistical analyses: A guide for non-statisticians. International Journal of Endocrinology and Metabolism, 10(2), 486-489.
  • Field, A. (2017). Discovering statistics using IBM SPSS statistics. Sage.
  • Wikipedia contributors. (2023). Skewness. https://en.wikipedia.org/wiki/Skewness

Related Assignments